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# EDO 9th Boyce Cap9 Resolução

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is at (3 , 2). The system is solved numerically, with y(0) = 2 and x(0) = 3.5 , 4.0 , 4.5 , 5.0 . The resulting periods are shown in the table: x(0) = 3.5 x(0) = 4.0 x(0) = 4.5 x(0) = 5.0 T 7.26 7.29 7.34 7.42 The actual amplitude steadily increases as the amplitude increases. 9. The system dx dt = a x(1\u2212 y 2 ) dy dt = b y(\u22121 + x 3 ) is solved numerically for various values of the parameters. The initial conditions are x(0) = 5 , y(0) = 2 . 500 Chapter 9. Nonlinear Differential Equations and Stability (a) a = 1 and b = 1 : The period is estimated by observing when the trajectory becomes a closed curve. In this case, T \u2248 6.45 . (b) a = 3 and a = 1/3 , with b = 1 : For a = 3 , T \u2248 3.69 . For a = 1/3 , T \u2248 11.44 . (c) b = 3 and b = 1/3 , with a = 1 : For b = 3 , T \u2248 3.82 . For b = 1/3 , T \u2248 11.06 . (d) It appears that if one of the parameters is fixed, the period varies inversely with 9.5 501 the other parameter. Hence one might postulate the relation T = k f(a , b) . 10.(a) Since T = 2pi/ \u221a ac , we first note that\u222b A+T A cos( \u221a ac t+ \u3c6)dt = \u222b A+T A sin( \u221a ac t+ \u3c6)dt = 0 . Hence x = 1 T \u222b A+T A c \u3b3 dt = c \u3b3 and y = 1 T \u222b A+T A a \u3b1 dt = a \u3b1 . (b) One way to estimate the mean values is to find a horizontal line such that the area above the line is approximately equal to the area under the line. From Figure 9.5.3, it appears that x \u2248 3.25 and y \u2248 2.0 . In Example 1 , a = 1 , c = 0.75 , \u3b1 = 0.5 and \u3b3 = 0.25 . Using the result in part (a), x = 3 and y = 2 . (c) The system dx dt = x(1\u2212 y 2 ) dy dt = y(\u22123 4 + x 4 ) is solved numerically for various initial conditions. x(0) = 3 and y(0) = 2.5 : 502 Chapter 9. Nonlinear Differential Equations and Stability x(0) = 3 and y(0) = 3.0 : x(0) = 3 and y(0) = 3.5 : x(0) = 3 and y(0) = 4.0 : It is evident that the mean values increase as the amplitude increases. That is, the mean values increase as the initial conditions move farther from the critical point. 9.5 503 12.(a) The equilibrium points are the solutions of the system x(a\u2212 \u3c3x\u2212 \u3b1y) = 0 y(\u2212c+ \u3b3x) = 0. If x = 0, then y = 0. If y = 0, then x = a/\u3c3. The third solution is found by substituting x = c/\u3b3 into the first equation. This implies that y = a/\u3b1\u2212 \u3c3c/(\u3b3\u3b1). So the equilibrium solutions are (0, 0), ( a\u3c3 , 0) and ( c \u3b3 , a \u3b1 \u2212 \u3c3c\u3b3\u3b1 ). When \u3c3 is increasing, the critical point ( a\u3c3 , 0) moves to the left and the critical point ( c \u3b3 , a \u3b1 \u2212 \u3c3c\u3b3\u3b1 ) moves down. The assumption a > \u3c3c\u3b3 is necessary for the third equilibrium to be in the first quadrant. (When a = \u3c3c\u3b3 , then the two nonzero critical points coincide.) (b) The Jacobian of the system is J = ( a\u2212 2\u3c3x\u2212 \u3b1y \u2212\u3b1x \u3b3y \u2212c+ \u3b3x ) . This implies that at the origin J(0, 0) = ( a 0 0 \u2212c ) , which implies that the origin is a saddle point. (a > 0 and c > 0 by our assumption.) At the critical point ( a\u3c3 , 0) J( a \u3c3 , 0) = (\u2212a \u2212\u3b1a/\u3c3 0 \u2212c+ \u3b3a/\u3c3 ) , which implies that this critical point is also a saddle as long as our assumption a > \u3c3c\u3b3 is valid. At the critical point ( c\u3b3 , a \u3b1 \u2212 \u3c3c\u3b3\u3b1 ) J( c \u3b3 , a \u3b1 \u2212 \u3c3c \u3b3\u3b1 ) = ( \u2212\u3c3c/\u3b3 \u2212\u3b1c/\u3b3 \u3b3a/\u3b1\u2212 \u3c3c/\u3b1 0 ) , which implies that this equilibrium point is either an asymptotically stable spiral or node. 13.(a) The equilibrium points are the solutions of the system x(1\u2212 x 5 \u2212 2y x+ 6 ) = 0 y(\u22121 4 + x x+ 6 ) = 0. If x = 0, then y = 0. If y = 0, then x = 5. The third equilibrium point can be found by setting 1/4 = x/(x+ 6), which gives x = 2 and then y = 2.4. So the critical points are (0, 0), (5, 0) and (2, 2.4). (b) The Jacobian of the system is J = ( 1\u2212 2x5 \u2212 12y(x+6)2 \u2212 2xx+6 6y (x+6)2 \u2212 14 + xx+6 ) . 504 Chapter 9. Nonlinear Differential Equations and Stability This implies that at the origin J(0, 0) = ( 1 0 0 \u22121/4 ) , which implies that the origin is a saddle point. At the critical point (5, 0) J(5, 0) = (\u22121 \u221210/11 0 9/44 ) , which implies that this critical point is also a saddle point. At the critical point (2, 2.4) J(2, 2.4) = (\u22121/4 \u22121/2 9/40 0 ) , which implies that this equilibrium point is an asymptotically stable spiral. 15.(a) Solving for the equilibrium of interest we obtain x = E2 + c \u3b3 y = a \u3b1 \u2212 \u3c3 \u3b1 · E2 + c \u3b3 \u2212 E1 \u3b1 . So if E1 > 0 and E2 = 0, then we have the same amount of prey and fewer predators. (b) If E1 = 0 and E2 > 0, then we have more prey and fewer predators. (c) If E1 > 0 and E2 > 0, then we have more prey and even fewer predators. 16.(b) The equilibrium solutions are given by the solutions of the system x(1\u2212 y 2 ) = H1 y(\u22123 4 + x 4 ) = H2. Now if H2 = 0, then x = 3 and the first equation gives y = 2\u2212 2H1/3. This means we have the same amount of prey and fewer predators. (c) If H1 = 0, then y = 2 and the second equation gives x = 3 + 2H2. This means we have the same amount of predators and more prey. 9.6 505 (d) If H1 > 0 and H2 > 0, then the second equation gives (x\u2212 3)y = 4H2 and using this we obtain from the first equation that x(1\u2212 2H2x\u22123 ) = H1. This gives the quadratic equation x2 \u2212 (3 + 2H2 +H1)x+ 3H1 = 0. Now at the old value x = 3 this expression is \u22126H2, so there is a root which is bigger than x = 3. The other root of the quadratic equation is closer to 0, so the equilibrium increases here: we have more prey. (Check this with e.g. H1 = H2 = 1: the roots are 3± \u221a 6, so both of the original roots get bigger.) A similar analysis shows that we will have fewer predators in this case. 9.6 2. We consider the function V (x, y) = a x2 + c y2 . The rate of change of V along any trajectory is V\u2d9 = Vx dx dt + Vy dy dt = 2ax(\u22121 2 x3 + 2xy2) + 2cy(\u2212y3) = = \u2212ax4 + 4ax2y2 \u2212 2cy4. Let us complete the square now the following way: V\u2d9 = \u2212ax4 + 4ax2y2 \u2212 2cy4 = \u2212a(x4 \u2212 4x2y2)\u2212 2cy4 = = \u2212a(x2 \u2212 2y2)2 + 4ay4 \u2212 2cy4 = \u2212a(x2 \u2212 2y2)2 + (4a\u2212 2c)y4. If a > 0 and c > 0 , then V (x, y) is positive definite. Clearly, if 4a\u2212 2c < 0, i.e. when c > 2a, then V\u2d9 (x, y) is negative definite. One such example is V (x, y) = x2 + 3 y2. It follows from Theorem 9.6.1 that the origin is an asymptotically stable critical point. 4. Given V (x, y) = a x2 + c y2 , the rate of change of V along any trajectory is V\u2d9 = Vx dx dt + Vy dy dt = 2ax(x3 \u2212 y3) + 2cy(2xy2 + 4x2y + 2y3) = = 2a x4 + (4c\u2212 2a)xy3 + 8c x2y2 + 4c y4. Setting a = 2c , V\u2d9 = 4c x4 + 8c x2y2 + 4c y4 \u2265 4c x4 + 4c y4. As long as a = 2c > 0 , the function V (x, y) is positive definite and V\u2d9 (x, y) is also positive definite. It follows from Theorem 9.6.2 that (0 , 0) is an unstable critical point. 5. Given V (x, y) = c(x2 + y2) , the rate of change of V along any trajectory is V\u2d9 = Vx dx dt + Vy dy dt = 2c x [y \u2212 xf(x, y)] + 2cy [\u2212x\u2212 yf(x, y)] = = \u22122c(x2 + y2)f(x, y) . 506 Chapter 9. Nonlinear Differential Equations and Stability If c > 0 , then V (x, y) is positive definite. Furthermore, if f(x, y) is positive in some neighborhood of the origin, then V\u2d9 (x, y) is negative definite. Theorem 9.6.1 asserts that the origin is an asymptotically stable critical point. On the other hand, if f(x, y) is negative in some neighborhood of the origin, then V (x, y) and V\u2d9 (x, y) are both positive definite. It follows from Theorem 9.6.2 that the origin is an unstable critical point. 9.(a) Letting x = u and y = u \u2032, we obtain the system of equations dx dt = y dy dt = \u2212g(x)\u2212 y . Since g(0) = 0 , it is evident that (0 , 0) is a critical point of the system. Consider the function V (x, y) = 1 2 y2 + \u222b x 0 g(s)ds . It is clear that V (0, 0) = 0 . Since g(u) is an odd function in a neighborhood of u = 0 , \u222b x 0 g(s)ds > 0 for x > 0 , and \u222b x 0 g(s)ds = \u2212 \u222b